按值传递二维数组不起作用 [英] Passing 2D array by value doesn't work

问题描述

我想传递一个二维数组到一个函数。我不希望实际数组修改。所以，按值传递。但我的实际2darray是越来越修改。（当我修改myArray的，X是越来越修改）。为什么会这样呢？

I am trying to pass a 2d array to a function. I don't want the actual array to be modified. So pass it by value. But my actual 2darray is getting modified.(when i modify myArray, x is getting modified). Why is it so?

int main(int argc, const char* argv[])
{
    int x[3][3];
    x[0][0]=2;
    x[0][1]=2;
    x[0][2]=2;
    x[1][0]=2;
    x[1][1]=2;
    x[1][2]=2;
    x[2][0]=2;
    x[2][1]=2;
    x[2][2]=2;
    func1(x);
}

void func1(int myArray[][3])
{
    int i, j;
    int ROWS=3;
    int COLS=3;

     for (i=0; i<ROWS; i++)
     {
         for (j=0; j<COLS; j++)
         {
             myArray[i][j] =3;
         }
     }
}

和有可能按值传递一个二维数组，如果我们不知道这两个方面？

And is it possible to pass a 2D array by value if we don't know both dimensions?.

推荐答案

要prevent func1的，将其更改为一个const指针。

to prevent func1 from being changed, change it to a const pointer.

void func1(const int (*myArray)[3])
{
    int i, j;
    int ROWS=3;
    int COLS=3;

     for (i=0; i<ROWS; i++)
     {
         for (j=0; j<COLS; j++)
         {
            // myArray[i][j] =3; //compilation error, is now const
            cout <<" "<<myArray[i][j];  
         }
     }
}

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